Re: [PHP-DB] Resource link errors

2002-02-09 Thread Ken Thompson



Thanks Rick, Now it works.


On Thursday 07 February 2002 03:04 pm, Rick Emery wrote:
> The following means you have not opened a link to your database.
> Concerntrate your efforts there.
>
> Warning: Supplied argument is not a valid MySQL-Link
> resource in /var/www/html/list.php3 on line 26
>
> -Original Message-
> From: Ken Thompson [mailto:[EMAIL PROTECTED]]
> Sent: Thursday, February 07, 2002 3:23 PM
> To: [EMAIL PROTECTED]
> Subject: [PHP-DB] Resource link errors
>
>
> Hello all,
> I'm very new to php and have been beating my head on the wall over this for
> 2
> days. I've searched through the php documentation but I guess I either
> don't
>
> know where or what to look for or don't understand what I'm seeing.
> I think this has to be an array but the original doesn't seem to think
> so.(see below)
> I've tried ($myrow = mysql_fetch_array($result)) and still get the same
> errors.
> The original that I used as an example works OK showing the results of
> Name,
>
> Position. They didn't include a data cell for $myrow[3] ...
> =
> THIS WORKS:
> 
> $title = "Welcome to the Web-Site of"; /*I commented this out and finaly
> removed it cuz I don't want the title echoed in the final page.*/
>
> include("header.inc");
>
> $result = mysql_query("SELECT * FROM employees",$db);
>
> echo "\n";
>
> echo "NamePosition\n";
>
> while ($myrow = mysql_fetch_row($result)) {
>
> echo("%s %s%s\n", $myrow[1], $myrow[2],
> $myrow[3]);
>
> }
>
> echo "\n";
> include("footer.inc");
>
> ?>
> ===
> What I am trying to do is to return the results of the query into a table
> with these items:
> +---+-+--+-+-++
>
> | Field | Type| Null | Key | Default | Extra  |
>
> +---+-+--+-+-++
>
> | id| tinyint(4)  |  | PRI | NULL| auto_increment |
> | year  | int(4)  | YES  | | NULL||
> | make  | varchar(20) | YES  | | NULL||
> | model | varchar(20) | YES  | | NULL||
> | engine| varchar(20) | YES  | | NULL||
> | fuel  | varchar(8)  | YES  | | NULL||
> | trans | varchar(10) | YES  | | NULL||
> | condition | varchar(20) | YES  | | NULL||
> | price | varchar(10) | YES  | | 0   ||
> | location  | varchar(20) | YES  | | NULL||
>
> +---+-+--+-+-++
> These are the errors that show up when I run the page in the browser.
> The database is local and I'm on Linux-Mandrake 8.1 using Apache web
> server. I can get the results I want from the command line, that is I can
> make queries and get the proper info returned.
> what have I done to cause it not to work?
>
>
> Warning: Supplied argument is not a valid MySQL-Link
> resource in /var/www/html/list.php3 on line 26
>
> Warning: Supplied argument is not a valid MySQL result
> resource in /var/www/html/list.php3 on line 32
>
>  THIS DOESN'T:
> I get the table and the item name e.g. Year, Make etc.
> Once I get the 3 cells returning the proper info, I want to add the rest of
> the cells needed to display all the info in the web page. I don't think it
> should be a problem, comments on this?
>
> 
> include("header.inc");
>
> //(Line 26 below, everything above " $result = mysql_query("SELECT * FROM autos",$db);
>
> echo "\n";
>
> echo "YearMakeModel\n";
>
> //(Line 32 below)
> while ($myrow = mysql_fetch_row($result)) {
>
> //What does the '%s' mean? I read it someplace in the tutorial
> //but can't find it now.
>
> printf("%s %s%s\n", $myrow[1], $myrow[2],
> $myrow[3]);
>
> }
>
> echo "\n";
> include("footer.inc");
>
> ?>

-- 

Ken Thompson, North West Antique Autos
Payette, Idaho
Email: [EMAIL PROTECTED]
http://www.nwaa.com
Sales and brokering of antique autos and parts.

Linux- Coming Soon To A Desktop Near You
Registered Linux User #183936


-- 
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php




RE: [PHP-DB] Resource link errors

2002-02-07 Thread Rick Emery

The following means you have not opened a link to your database.
Concerntrate your efforts there.

Warning: Supplied argument is not a valid MySQL-Link 
resource in /var/www/html/list.php3 on line 26

-Original Message-
From: Ken Thompson [mailto:[EMAIL PROTECTED]]
Sent: Thursday, February 07, 2002 3:23 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Resource link errors


Hello all,
I'm very new to php and have been beating my head on the wall over this for
2 
days. I've searched through the php documentation but I guess I either don't

know where or what to look for or don't understand what I'm seeing.
I think this has to be an array but the original doesn't seem to think
so.(see below) 
I've tried ($myrow = mysql_fetch_array($result)) and still get the same 
errors.
The original that I used as an example works OK showing the results of Name,

Position. They didn't include a data cell for $myrow[3] ...
=
THIS WORKS:
\n";

echo "NamePosition\n";

while ($myrow = mysql_fetch_row($result)) {

echo("%s %s%s\n", $myrow[1], $myrow[2], 
$myrow[3]);

}

echo "\n";
include("footer.inc");

?>
===
What I am trying to do is to return the results of the query into a table 
with these items:
+---+-+--+-+-++
| Field | Type| Null | Key | Default | Extra  |
+---+-+--+-+-++
| id| tinyint(4)  |  | PRI | NULL| auto_increment |
| year  | int(4)  | YES  | | NULL||
| make  | varchar(20) | YES  | | NULL||
| model | varchar(20) | YES  | | NULL||
| engine| varchar(20) | YES  | | NULL||
| fuel  | varchar(8)  | YES  | | NULL||
| trans | varchar(10) | YES  | | NULL||
| condition | varchar(20) | YES  | | NULL||
| price | varchar(10) | YES  | | 0   ||
| location  | varchar(20) | YES  | | NULL||
+---+-+--+-+-++
These are the errors that show up when I run the page in the browser.
The database is local and I'm on Linux-Mandrake 8.1 using Apache web server.
I can get the results I want from the command line, that is I can make 
queries and get the proper info returned. 
what have I done to cause it not to work?


Warning: Supplied argument is not a valid MySQL-Link 
resource in /var/www/html/list.php3 on line 26
 
Warning: Supplied argument is not a valid MySQL result 
resource in /var/www/html/list.php3 on line 32

 THIS DOESN'T:
I get the table and the item name e.g. Year, Make etc. 
Once I get the 3 cells returning the proper info, I want to add the rest of 
the cells needed to display all the info in the web page. I don't think it 
should be a problem, comments on this?

\n";

echo "YearMakeModel\n";

//(Line 32 below) 
while ($myrow = mysql_fetch_row($result)) {

//What does the '%s' mean? I read it someplace in the tutorial
//but can't find it now.

printf("%s %s%s\n", $myrow[1], $myrow[2], 
$myrow[3]);

}

echo "\n";
include("footer.inc");

?>
-- 

Ken Thompson, North West Antique Autos
Payette, Idaho
Email: [EMAIL PROTECTED]
http://www.nwaa.com
Sales and brokering of antique autos and parts.

Linux- Coming Soon To A Desktop Near You
Registered Linux User #183936

-- 
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php

-- 
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php